JEE Mains · Physics · STD 12 - 5. Magnetism and matter
A short bar magnet placed with its axis at \(30^{\circ}\) with an external field of 800 Gauss, experiences a torque of \(0.016 N . m\). The work done in moving it from most stable to most unstable position is \(\alpha \times 10^{-3} J\). The value of \(\alpha\) is _________ .
- A 32
- B 64
- C 16
- D 128
Answer & Solution
Correct Answer
(B) 64
Step-by-step Solution
Detailed explanation
\( \tau=\mu B~sin~\theta\Rightarrow0.016=\mu\times B\times\frac{1}{2} \) \( \Rightarrow\mu=\frac{0.032}{B} \) \( W_{ext}=U_{f}-U_{i}=\mu B-(-\mu B)=2\mu B \) \( =2\times\frac{0.032}{B}\times B \) \( =0.064~J \)
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